22 Coupled Antennas22.1 Near Fields of Linear Antennas In calculating mutual coupling effects between closely-spaced linear antennas, we need to know the fields produced by an antenna at
Trang 122 Coupled Antennas
22.1 Near Fields of Linear Antennas
In calculating mutual coupling effects between closely-spaced linear antennas, we need
to know the fields produced by an antenna at near distances The fields generated by a
thin wire antenna with currentI(z)were worked out in Sec 14.4
We summarize these results here All field components can be obtained from the
knowledge of thez-component of the magnetic vector potentialAz(z, ρ):
whereh is the half-length of the antenna, h = l/2, and the geometry is shown in
Fig 22.1.1 We have used the approximate thin-wire kernel because it differs little from
the exact kernel for distancesρ > a(typically, whenρ5a.)
Fig 22.1.1 Fields of a thin wire antenna.
(22.1.2)
As a first approximation, we will assume that the currentI(z)is sinusoidal This
is justified only when the antenna length is near half a wavelengthλ/2 Most coupledantenna arrays that are used in practice, such as Yagi-Uda, satisfy this condition
We note also that the near fields resulting from the sinusoidal current assumption
do not satisfy the correct boundary conditions on the surface of the antenna, that isthe conditionEz(z, ρ)=0 atz=0 andρ= a In Sec 22.2, we consider an improvedapproximation of the near fields that addresses these issues Thus for now, we willassume that:
I(z)= I0
sink(h− |z|)sinkh = Imsin
k(h− |z|) (22.1.3)where we distinguish between the currentI0atz=0 and the maximum currentIm=
I0/sinkh For half-wavelength antennas, we havekh= π/2,I0= Im, and the currentbecomesI(z)= I0coskz
In principle, one could insert Eq (22.1.3) into (22.1.1) and perform the requiredintegrations to getAz However, for the purpose of determining the fields, this is notnecessary Combining (22.1.1) and (22.1.2), we obtain:
k2)I(z)=0, and therefore, the integrand of (22.1.4) becomes a complete derivative:I(z)(∂2z+ k2)G(z− z, ρ)= ∂z
I(z)∂z G(z− z, ρ)−G(z − z, ρ)∂
z I(z)
(22.1.6)Integrating the first term, we obtain:
h
−h∂z
I(z)∂z G(z− z, ρ)
dz= I(h)∂z G(z− h, ρ)−I(−h)∂z G(z+ h, ρ)=0where we used the end-conditionsI(h)= I(−h)=0 The second term in (22.1.6) is alittle trickier because∂zI(z)is discontinuous atz=0 Splitting the integration range,
we obtain:
h
−h∂z
G(z− z, ρ)∂zI(z)
dz=
0
−h+
h 0
∂z G(z− z, ρ)∂zI(z)
dz
=G(z, ρ)I(0−)−G(z + h, ρ)I(−h)+G(z− h, ρ)I(h)−G(z, ρ)I(0+)
= kI 2 coskh G(z, ρ)−G(z − h, ρ)−G(z + h, ρ)
Trang 222.1 Near Fields of Linear Antennas 907
where we usedI(0±)= ∓kImcoskhandI(±h)= ∓kIm Inserting this result into
Eq (22.1.4) and rearranging some constants, we find:
Ez(z, ρ)= −jηIm
4π
G(z− h, ρ)+G(z + h, ρ)−2 coskh G(z, ρ)
(22.1.7)The quantitiesG(z− h, ρ), G(z + h, ρ), G(z, ρ)can be written conveniently as follows:
G(z, ρ)=e−jkR0
R0, R0=ρ2+ z2
G(z− h, ρ) =e−jkR1
R1, R1=ρ2+ (z − h)2
G(z+ h, ρ) =e−jkR2
R2, R2=ρ2+ (z + h)2
(22.1.8)
whereR0, R1, R2are recognized to be the distances from the center and the two ends
of the antenna to the observation point, as shown in Fig 22.1.1 Thus, we can write:
Next, we determineHφfrom Amp`ere’s law in (22.1.2) by noting thatρEzis a complete
derivative with respect toρ Indeed, for any of the quantitiesR, we have:
A possible integration constant inρis dropped because the field must vanish when
its source vanishes, that is, whenIm=0 Finally, we obtainEρfrom Faraday’s law in
(22.1.2) Noting the differentiation property:
pur-It is worth also to verify that the exact expressions for the fields give correctly theradiation fields that were derived in Sec 16.3 At large distances, we can make theapproximations:
R0= r, R1= r − hcosθ, R2= r + hcosθwhereris the radial distance andθthe polar angle Replacingρ= rsinθ, the magneticfield (22.1.10) becomes approximately:
This agrees with the results of Sec 16.3
22.2 Improved Near-Field Calculation
The current on a thin linear antenna is determined from the solution of the Hall´en orPocklington integral equations; for example, the latter is,
As we saw in Sec 21.4, the assumption of a sinusoidal current can be justified on thebasis of Pocklington’s equation, but it represents at best a crude approximation Theresulting electric field does not satisfy condition (22.2.2), as can be seen settingρ= ainto Eq (22.1.9)
King’s three-term approximation, or a three-term fitted to a numerical solution, vides a better approximation to the current, and one may expect that the fields generated
Trang 3pro-22.2 Improved Near-Field Calculation 909
by such current would more closely satisfy the boundary condition (22.2.2) This is what
we discuss in this section
Because the current need not satisfy the Helmholtz equation,I(z)+k2I(z)=0, we
must revisit the calculations of the previous section We begin by assuming thatI(z)is
symmetric inzand that it vanishes at the antenna end-points, that is,I(±h)=0 The
electric fieldEz(z, ρ)at distanceρis obtained from Eq (22.1.4):
=h
z =−h
(22.2.4)
The assumed symmetry ofI(z)implies a discontinuity of its derivative atz=0
In-deed, settingI(z)= F(|z|), for some continuous and continuously differentiable
0 +
0 −G(z− z, ρ)
I(z)+k2I(z)
dz=2G(z, ρ)F(0)
G(z− z, ρ)I(z) 0+
0 −=2G(z, ρ)F(0)Using the following notation for the principal-value integral,
I(z)+k2I(z)
dz++2I(0+)e−jkR0
e−jkR1+ e−jkR 2 (22.2.7)which may also be written in the form:
I(z)= A1
sin(k|z|)−sin(kh)
+A2
cos(kz)−cos(kh)
+A3
cos
kz2
−cos
kh2(22.2.11)and fix the coefficientsA1, A2, A3by fitting this expression to a numerical solution asdiscussed in Sec 21.6, and then, use Eq (22.2.11) into (22.2.6) with the integral term
Trang 422.2 Improved Near-Field Calculation 911
1 2 3 4
20 40 60 80
z/λ
imaginary part of E
z (z,a)
total correction term standard term
Fig 22.2.1 Calculated near fieldEz(z, ρ)forl=0.5λ
evaluated numerically Fig 22.2.1 shows the results of such a calculation for a
half-wave antennal =0.5λwith radiusa = 0.005λ Fig 22.2.2 shows the results for a
full-wave antennal=1.0λwith the same radius The required quantities appearing in
(22.2.6) are calculated as follows:
I(z)+k2I(z)= −k2A1sinkh− k2A2coskh− k2A3
cos
kh2
0 0.1 0.2 0.3 0.4 0.5 0
0.5 1 1.5 2 2.5
1 2
−2
−1 0 1 2
ln(ρ/a)
l = 0.5 λ, a = 0.005λ, z = 0.2h
real part real part log(ρ) approx imag part imag part log( ρ) approx
0 0.1 0.2 0.3 0.4 0.5 0
10 20 30 40
z/λ
real part of E
z (z,a)
total correction term standard term
0 0.1 0.2 0.3 0.4 0.5
−40
−20 0 20
z/λ
imaginary part of E
z (z,a)
total correction term standard term
Fig 22.2.2 Calculated near fieldEz(z, ρ)forl=1.0λ
I(0+)= −I(0−)= kA1
I(h)= −I(−h)= kA1coskh− kA2sinkh−1
2kA3sin
kh2
The numerical solutions were obtained by solving the Hall´en equation with matching, pulse basis functions, and the exact kernel usingM=100 upper-half current
Trang 5point-22.2 Improved Near-Field Calculation 913
samplesIn These current samples were then used as in Eq (21.6.15) to obtain the
parametersA1, A2, A3
The upper-left graphs show the currentI(z)of Eq (22.2.11) together with the
sam-plesInto which it was fitted
The upper-right graphs show the magnitude ofEz(z, ρ)as a function ofρforzfixed
atz=0.2h The behavior ofEz(z, ρ)is consistent initially with a logarithmic
depen-dence onρas predicted by King and Wu [1238,1239] and discussed below, followed then
by the expected 1/ρdecrease arising from the last three standard terms of Eq (22.2.6),
which are represented by the dashed curves
The left middle-row graphs display the logarithmic dependence more clearly by
plot-ting the real and imaginary parts ofEz(z, ρ)versus ln(ρ/a), including the King-Wu
approximation of Eq (22.2.15)
The right middle-row graphs show the magnitude of the fieldEz(z, a)at the surface
of the antenna as a function ofzover the interval 0≤ z ≤ h Except at the feed and end
points, the field is effectively zero as required by the boundary conditions
To observe the importance of the correction term, that is, the principal-value integral
in Eq (22.2.6), the third-row graphs display the real and imaginary parts ofEz(z, a)
versusz Plotted separately are also the correction and standard terms, which appear
always to have opposite signs canceling each other so that the net field is zero
The graphs for Fig 22.2.1 were generated by the following MATLAB code (for Figure
22.2.2 simply setL=1):
L = 0.5; h = L/2; a = 0.005; k = 2*pi; eta = 377;
z = 0:h/100:h;
Helm = @(z) -k^2*(A(1)*sin(k*h) + A(2)*cos(k*h) + A(3)*(cos(k*h/2)-3/4*cos(k*z/2)));
z = 0.2*h; r = linspace(a,200*a, 1001); logr = log(r/a);
for i=1:length(r),
GHelm = G(z-zi,r(i)) * Helm(zi);
E2(i) = (- I1h * (G(z-h,r(i)) + G(z+h,r(i))) + 2*I10 * G(z,r(i))) * S;
E(i) = E1(i) + E2(i);
end
figure; plot(logr,real(E), logr,real(Eapp),’ ’,
clear E E1 E2;
z = linspace(0,h,201); r = a;
for i=1:length(z), GHelm = G(z(i)-zi,r) * Helm(zi);
E1(i) = (wi’*GHelm) * S;
E2(i) = (- I1h * (G(z(i)-h,r) + G(z(i)+h,r)) + 2*I10 * G(z(i),r)) * S;
E(i) = E1(i) + E2(i);
end
figure; plot(z,real(E), z,real(E1),’ ’, z,real(E2),’:’); % lower-left graph figure; plot(z,imag(E), z,imag(E1),’ ’, z,imag(E2),’:’); % lower-right graphNext, we discuss the King-Wu small-ρapproximation [1238,1239]; see also McDonald[1293] First, we note that theHφ andEρ components in Eqs (22.2.8) and (22.2.10)were obtained by using Maxwell’s equations (22.1.2), that is, Amp`ere’s laws∂ρ(ρHφ)=jω ρEzandjωEρ= −∂zHφ We may also verify Faraday’s law, which has only aφcomponent in this case:
∂ρEz− ∂zEρ= jωμHφ (22.2.12)Indeed, this can be derived from Eqs (22.2.3), (22.2.8), and (22.2.10) by using the identity:
ρ ∂
∂ρ
e−jkRR
+ ∂
ρ→ 0 in the right-hand side of Eq (22.2.8) In this limit, we havee−jkR= e−jk|z−z |,which is recognized as the Green’s function of the one-dimensional Helmholtz equationdiscussed in Sec 21.3 that satisfies(∂2
jωEρ= −∂zHφ= −I2(z)
πρ =jωQ(z)2
πρthat is, for smallρ:
Trang 622.2 Improved Near-Field Calculation 915
The same result can also be derived from Eq (22.2.10) by recognizing the small-ρ
limit(z− z)e−jkR/R→sign(z− z)e−jk|z−z |, which satisfies the Helmholtz identity:
(∂2z+ k2
)sign(z− z)e−jk|z−z |=2∂zδ(z− z)Combining Eqs (22.2.13) and (22.2.14) into the Faraday equation (22.2.12), we have,
∂ρEz= ∂zEρ+ jωμHφ=Q2(z)
πρ + jωμI(z)2
πρ = jω
I(z)+k2I(z)
2πρIntegrating fromρ= a, we obtain the small-ρKing-Wu approximation:
Ez(z, ρ)= Ez(z, a)+2πωj I(z)+k2
I(z)ln
ρa
(22.2.15)Strictly speaking, we must setEz(z, a)=0 because of the boundary condition How-
ever, in our numerical solution, we have kept the termEz(z, a), which is small but not
necessarily exactly zero, in order to compare the analytical calculation (22.2.15) with
the numerical solution The left middle-row graphs confirm the linear dependence on
ln(ρ/a)with the right slope
For longer antennas, up to aboutl=3λ, the four-term approximation discussed in
Sec 21.6 can be used and leads to similar results In this case, the following current
expressions should be used:
I(z)= A1
sin(k|z|)−sin(kh)
+ A2
cos(kz)−cos(kh)
++ A3
cos
kz4
3kz4
−cos
3kh4
I(z)+k2I(z)= −k2A1sinkh− k2A2coskh− k2A3
cos
kh4
−1516cos
kz4
− k2A4
cos
3kh4
−167 cos
3
kz4
I(0+)= −I(0−)= kA1
I(h)= −I(−h)= kA1coskh− kA2sinkh−14kA3sin
kh4
−34kA4sin
3kh4
and this remains roughly true for antenna lengths 0.5≤ l/λ ≤1.3 and radii 0.001≤
a/λ≤0.007 and for a variety of distances along the antenna, such as, 0.2h≤ z ≤0.7h
Thus, this distance may be taken as a rough measure of the distance beyond which
the standard terms begin to take over and the sinusoidal current approximation becomes
justified
The mutual impedance formulas that we develop in succeeding sections are based
on the sinusoidal assumption, and therefore, they can be used more reliably for antenna
separationsdthat are greater than that of Eq (22.2.16) For example, to increase one’s
confidence, one could take the separations to be greater than, say, double the above
value, that is,d≥ λ/10
22.3 Self and Mutual Impedance
The mutual coupling between antennas cannot be ignored if the antennas are near eachother The mutual impedance is a measure of such proximity effects [2,1308–1320].Consider two parallel center-driven linear dipoles, as shown in Fig 22.3.1 Theirdistance along thex-direction isdand their centers are offset bybalong thez-direction
Fig 22.3.1 Parallel linear dipoles.
If antenna-1 is driven and antenna-2 is open-circuited, the near field generated bythe current on antenna-1 will cause an open-circuit voltage, sayV21,ocon antenna-2 Themutual impedance of antenna-2 due to antenna-1 is defined to be:
Z21=V21,oc
I1
(22.3.1)whereI1is the input current on antenna-1 Reciprocity implies thatZ12= Z21 Moregenerally, if both antennas are driven, then, the relationship of the driving voltages tothe input currents is given by:
approx-In order to derive convenient expressions that allow the calculation of the mutualand self impedances, we use the reciprocity result given in Eq (21.5.6) for the short-circuit current and open-circuit voltage induced on a receiving antenna in the presence
of an incident field
If antenna-2 is open-circuited and thez-component of the electric field generated
by antenna-1 and incident on antenna-2 isE21(z), then according to Eq (21.5.6), theinduced open-circuit voltage will be:
V21,oc= −1
I
h 2
Trang 722.3 Self and Mutual Impedance 917
whereh2= l2/2, andI2(z),I2= I2(0)are the current and input current on antenna-2
when it is transmitting It follows from definition (22.3.1) that:
I1(z)= I1
sink(h1− |z|)sinkh1 = Im1sin
k(h1− |z|)
I2(z)= I2
sink(h2− |z|)sinkh2 = Im2sin
k(h2− |z|)then, according to Eq (22.1.9) the electric fieldE21(z)along antenna-2 will be:
Inserting Eq (22.3.5) into (22.3.4) and rearranging some constants, we find the final
expression for the mutual impedanceZ21:
one or both of the antennas have lengths that are multiples ofλ, then one or both of
the denominator factors sinkh1, sinkh2will vanish resulting in an infinite value for the
mutual impedance
This limitation is caused by the sinusoidal current assumption We saw in Chap 21
that the actual input currents are not zero in a real antenna On the other hand, in most
applications of Eq (22.3.7) the lengths differ slightly from half-wavelength for which the
sinusoidal approximation is good
The definition (22.3.4) can also be referred to the maximum currents by normalizing
by the factorIm1Im2, instead ofI1I2 In this case, the mutual impedance isZ21m =
Z21sinkh1sinkh2, that is,
R0=a2+ z2, R1=a2+ (z − h1)2, R2=a2+ (z + h1)2 (22.3.12)The MATLAB functionimped implements Eq (22.3.7), as well as (22.3.10) It returnsbothZ21andZ21mand has usage:
[Z21,Z21m] = imped(L2,L1,d,b) % mutual impedance of dipole 2 due to dipole 1
[Z21,Z21m] = imped(L2,L1,d) % b = 0, side-by-side arrangement
[Z,Zm] = imped(L,a) % self impedancewhere all the lengths are in units ofλ The function uses 16-point Gauss-Legendreintegration, implemented with the help of the functionquadr, to perform the integral
in Eq (22.3.7)
In evaluating the self impedance of an antenna with a small radius, the integrandF(z)varies rapidly aroundz=0 To maintain accuracy in the integration, we split theintegration interval into three subintervals, as we mentioned in Sec 21.10
Gauss-Legendre integration, the integrandF(z)is never evaluated atz=0, even if the antennaradius is zero This allows us to estimate the self-impedance of an infinitely thin half-wavelength antenna by settingL=0.5 anda=0:
Z=imped(0.5,0)=73.0790+42.5151j ΩSimilarly, for radiia=0.001λand 0.005λ, we find:
Z=imped(0.5,0.001)=73.0784+42.2107j Ω
Z=imped(0.5,0.005)=73.0642+40.6319j Ω
A resonant antenna is obtained by adjusting the lengthLsuch that the reactance part ofZbecomes zero The resonant length depends on the antenna radius For zero radius, thislength isL=0.48574823 and the corresponding impedance,Z=67.1843 Ω
arrange-ment separated by distanced The antenna radius isa=0.001 and therefore, its selfimpedance is as in the previous example If antenna-1 is driven and antenna-2 is parasitic,that is, short-circuited, then Eq (22.3.2) gives:
V1= Z11I1+ Z12I2
0= Z I + Z I
Trang 822.3 Self and Mutual Impedance 919
Solving the second for the parasitic currentI2= −I1Z21/Z22and substituting in the first,
we obtain driving-point impedance of the first antenna:
Z2 11
−40 0 40
Fig 22.3.2 Mutual impedance between identical half-wave dipoles vs separation.
For separationsdthat are much larger than the antenna lengths, the impedanceZ21
falls like 1/d Indeed, it follows from Eq (22.3.6) that for larged, all three distances
R0, R1, R2become equal tod Therefore, (22.3.8) tends to:
Z21→ jη(1−coskh1)(1−coskh2)
πsinkh1sinkh2
e−jkd
The envelope of this asymptotic form was superimposed on the graph of Fig 22.3.2
The oscillatory behavior ofZ21with distance is essentially due to the factore−jkd
An alternative computation method of the mutual impedance is to reduce the
inte-grals (22.3.7) to the exponential integralE1(z)defined in Appendix F, taking advantage
of MATLAB’s built-in functionexpint
By folding the integration range[−h1, h1]in half and writing sin
k(h2− |z|)as asum of exponentials, Eq (22.3.7) can be reduced to a sum of terms of the form:
G(z0, s)=
h 1 0
e−jkR
−jkszdz , R=d2+ (z − z0)2, s= ±1 (22.3.14)which can be evaluated in terms ofE1(z)as:
G(z0, s)= se−jksz 0
E1(ju0)−E1(ju1)
(22.3.15)with
u0= k
d2+ z2− sz0
u1= k
d2+ (h1− z0)2+ s(h1− z0)Indeed, the integral in (22.3.7) can be written as a linear combination of 10 such terms:
E1(ju)= −γ −lnu+ Cin(u)+jSi(u)−π2
as defined in Eq (F.27), then (22.3.10) can be expressed in the following form, where wesetZ11= Zin= Rin+ jXin,h1= h, andl=2h:
2Cin(kl+)−Cin(kL+)+2Cin(kl−)−Cin(kL−)−2Cin(ka)+12sinkl
2Si(kl−)−Si(kL−)+Si(kL+)−2Si(kl+)
(22.3.18)
Trang 922.3 Self and Mutual Impedance 921
These expressions simplify substantially if we assume that the radiusais small, as
is the case in practice In particular, assuming thatka 1 anda , the quantities
l±andL±can be approximated by:
l+2h= l , l−=a2
l+ a2l
L+4h=2l , L−= a2
L+a2
2l
(22.3.20)
Noting thatSi(x)andCin(x)vanish atx=0, we may neglect all the terms whose
arguments arekl−,kL−, orka, and replacekl+= klandkL+=2kl, obtaining:
We note thatAis independent of the radiusaand leads to the same expression for
the radiation resistance that we found in Sec 16.3 using Poynting methods
An additional approximation can be made for the case of a small dipole Assuming
thatkh 1, in addition toka 1 anda , we may expand each of the above terms
into a Taylor series in the variablekhusing the following Taylor series expansions of
the functionsSi(x)andCin(x):
whereL=ln(2a/l) The resistanceRis identical to that obtained using the Poynting
method and assuming a linear approximation to the sinusoidal antenna current, which
is justified whenkh 1:
I(z)= I0
sink(h− |z|)sinkh I0
22.4 Coupled Two-Element Arrays
Next, we consider a more precise justification of Eq (22.3.2) and generalize it to thecase of an arbitrary array of parallel linear antennas Fig 22.4.1 shows twoz-directedparallel dipoles with centers at locations(x1, y1)and(x2, y2)
We assume that the dipoles are center-driven by the voltage generatorsV1, V2 Let
I1(z), I2(z)be the currents induced on the dipoles by the generators and by their tual interaction, and leth1, h2 be the half-lengths of the antennas, and a1, a2, theirradii Then, assuming the thin-wire model, the total current density will have only a
mu-z-component given by:
Jz(x, y, z)= I1(z)δ(x− x1)δ(y− y1)+I2(z)δ(x− x2)δ(y− y2) (22.4.1)
Fig 22.4.1 Array of two linear antennas.
It follows that the magnetic vector potential will be:
Az(z, ρρρ)=4μπ
e−jkR
R Jz(x, y, z)dxdydz R= |r−r|whereρ= xˆx+ yy is the cylindrical radial vector Inserting (22.4.1) and performingˆthex, yintegrations, we obtain:
to the(x, y, z)observation point, that is,
R1=(z− z)2+(x − x1)2+(y − y1)2=(z− z)2+|ρρρ −d1|2
R2=(z− z)2+(x − x2)2+(y − y2)2=(z− z)2+|ρρρ −d2|2
(22.4.3)
where d1= (x1, y1)and d2= (x2, y2)are thexy-locations of the antenna centers The
z-component of the electric field generated by the two antenna currents will be:
jωμ E (z, ρρρ)= (∂2+ k2)A (z, ρρρ)
Trang 1022.4 Coupled Two-Element Arrays 923
Working with the rescaled vector potentialV(z, ρρρ)=2jcAz(z, ρρρ), we rewrite:
and antenna-2, we obtain from Eq (22.4.4):
V1(z)= V11(z)+V12(z)
Thez-components of the electric fields induced on the surfaces of antenna-1 and
antenna-2 are obtained by applying Eq (22.4.5) to each term of (22.4.6):
Ifp= q, thendpqis thexy-distance between the antennas, and ifp= q, it is the
radius of the corresponding antenna, that is,
d12= d21= |d1−d2| =(x1− x2)2+(y1− y2)2
d11= a1, d22= a2
(22.4.10)
Thus,Vpq(z)andEpq(z)are the vector potential and thez-component of the electric
field induced on antenna-pby the currentIq(z)on antenna-q
To clarify these definitions, Fig 22.4.2 shows a projected view of Fig 22.4.1 on the
xyplane The pointPwith radial vectorρis the projection of the observation point
(z, ρρρ) WhenPcoincides with a point, such asP2, on the surface of antenna-2 defined
by the radial vectorρ2, then the distance(P2O2)= |ρρρ2−d2|will be equal to the antenna
radiusa2, regardless of the location ofP2around the periphery of the antenna
On the other hand, the distance(P2O1)= |ρρρ2−d1|varies withP2 However, because
the separationd12is typicallyd12 a2, such variation is minor and we may replace
|ρρρ2−d1|by|d2−d1| Thus, in evaluatingV(z, ρρ2)on antenna-2, we may use Eq (22.4.4)
withR , R defined by:
Fig 22.4.2 Array of two linear antennas.
Now, on the surface of the first antenna, the electric fieldEzmust cancel the field ofthe delta-gap generator in order for the total tangential field to vanish, that is,E1(z)=
−E1,in(z)= −V1δ(z) Similarly, on the surface of the second antenna, we must have
E2(z)= −E2,in(z)= −V2δ(z) Then, Eq (22.4.7) becomes:
E11(z)+E12(z)= E1(z)= −V1δ(z)
E21(z)+E22(z)= E2(z)= −V2δ(z) (22.4.12)Combining these with the Eq (22.4.8), we obtain the coupled version of the Hall´en-Pocklington equations:
(∂2
z+ k2)
V11(z)+V12(z)
=2kV1δ(z)(∂2
Trang 1122.5 Arrays of Parallel Dipoles 925
Using these definitions and Eq (22.4.12), we find:
The mutual impedance defined in Eq (22.4.14) actually satisfies the reciprocity
sym-metry condition,Zpq= Zqp To write it in a form that shows this condition explicitly,
we replaceEpq(z)by (22.4.8) and obtain the alternative symmetric form:
z+ k2)Gpq(z− z)dz dz (22.4.15)
If we assume that the currents are sinusoidal, that is, forp=1,2,
Ip(z)= Ip
sink(hp− |z|)sinkhp
(22.4.16)
then, in Eq (22.4.15) the ratiosIp(z)/Ipand henceZpqbecome independent of the input
currents at the antenna terminals and depend only on the geometry of the antennas
22.5 Arrays of Parallel Dipoles
The above results on two antennas generalize in a straightforward fashion to several
antennas Fig 22.5.1 depicts the case ofKparallel dipoles in side-by-side arrangement
with centers at positions(xp, yp), and driving voltages, lengths, half-lengths, and radii,
Vp, lp, hp, ap, wherep=1,2 , K
Fig 22.5.1 Two-dimensional array of parallel dipoles.
Assuming sinusoidal currents as in Eq (22.4.16), we define the mutual impedances
Zpqby Eq (22.4.14) or (22.4.15), wherep, qtake on the valuesp, q=1,2 , K The
Hall´en-Pocklington equations (22.4.13) generalize into:
(∂2
z+ k2)K
q =1
ZpqIq , p=1,2, , K (22.5.3)
whereIqis the input current at the center of theqth antenna Eq (22.5.3) may be written
in a compact matrix form:
... antenna-2 is open-circuited and thez-component of the electric field generatedby antenna-1 and incident on antenna-2 isE21(z), then according to Eq (21.5.6), theinduced open-circuit...
If antenna-1 is driven and antenna-2 is open-circuited, the near field generated bythe current on antenna-1 will cause an open-circuit voltage, sayV21,ocon antenna-2 Themutual... functions, and the exact kernel usingM=100 upper-half current
Trang 5point -2 2. 2 Improved Near-Field